inverse laplace (2s+1)/(s^2+s) — step-by-step solution

⇥1) You can decompose into partial fractions:

⇥

\frac{2 s + 1}{s \cdot (s + 1)} = \frac{A}{s} + \frac{B}{s + 1}

⇥For the simple factor:

s

and its root:

0

, compute A by multiplying by the denominator:

s \cdot (s + 1)

and evaluating at the root (the other terms cancel out):

⇥2) Let's solve in

ℝ

(0 + 1) \cdot A = 2 \cdot 0 + 1

⇥ⓘ Adding 0 changes nothing:

0 + 1

1

■A product with a zero factor is zero:

2 \cdot 0

0

⇥

A = 0 + 1

⇥ⓘ Adding 0 changes nothing:

0 + 1

1

⇥Solution found:

A = 1

⇥For the simple factor:

s + 1

and its root:

-1

, compute B by multiplying by the denominator:

s \cdot (s + 1)

and evaluating at the root (the other terms cancel out):

⇥3) Let's solve in

ℝ

-1 \cdot B = 2 \cdot (-1) + 1

⇥ⓘ Multiplying by -1 gives the opposite:

-1 \cdot B

- B

■Multiplications are done before additions and subtractions. A single minus sign does not cancel out, the product of a negative number by a positive number is negative:

2 \cdot (-1)

-2

⇥

- B = - 2 + 1

⇥ⓘ Start from

-2

and move forward

1

on the number line, getting closer to zero:

- 2 + 1

-1

⇥

- B = -1

⇥Solution found:

B = 1

⇥4) Finally, you can substitute the values of the unknowns into the initial partial fraction decomposition:

⇥

\frac{2 s + 1}{s \cdot (s + 1)} = \frac{1}{s} + \frac{1}{s + 1}

Find the causal inverse transform, for t ≥ 0. By linearity, invert each partial fraction

The inverse Laplace formula for

\frac{1}{{(s - a)}^{n}}

t^{n - 1} \cdot \frac{\exp (a t)}{(n - 1)!}

; it gives for

\frac{1}{s}

\frac{\exp (0 \cdot t)}{0!}

The inverse Laplace formula for

\frac{1}{{(s - a)}^{n}}

t^{n - 1} \cdot \frac{\exp (a t)}{(n - 1)!}

; it gives for

\frac{1}{s + 1}

\frac{\exp (-1 \cdot t)}{0!}

⇥

\frac{\exp (0 \cdot t)}{0!} + \frac{\exp (-1 \cdot t)}{0!}

⇥ⓘ A product with a zero factor is zero:

0 \cdot t

0

■Any nonzero number raised to the power 0 is 1:

\exp (0)

1

■By convention, the factorial of 0 is 1:

0!

1

■Multiplying by -1 gives the opposite:

-1 \cdot t

- t

⇥

\frac{1}{1} + \frac{\exp (- t)}{1}

⇥ⓘ The factor is moved to the numerator or the denominator by changing the sign of its exponent, for a nonzero base:

\exp (- t)

\frac{1}{\exp (t)}

⇥

1 + \frac{1}{\exp (t)}

▶Inverse Laplace transform:

\frac{1}{\exp (t)} + 1